End of chapter exercises

We first find the gradient of the line:

\begin{align*} m_{AB} & = \frac{y_2-y_1}{x_2-x_1}\\ & = \frac{1 - 3}{-6 - 0}\\ & = \frac{-2}{-6}\\ & = \frac{1}{3} \end{align*}

Next we note that point \(B\) lies on the \(y\)-axis and so is the \(y\)-intercept. Therefore \(c = 3\).

Therefore the equation of the line \(AB\) is: \(y = \frac{1}{3}x+3\).

\begin{align*} d_{AB} &= \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\ &= \sqrt{(-6 + 0)^2 + (1 - 3)^2}\\ &= \sqrt{(-6)^2 + (-2)^2}\\ &= \sqrt{40} \end{align*}

\begin{align*} m &= \frac{y_2-y_1}{x_2-x_1} \\ m_{PQ} &= \frac{12-5}{-2-5} \\ &= \frac{7}{-7} \\ m_{PQ} &= -1 \\ \\ y &= mx + c \\ (5) &= (-1)(5) + c \\ c &= 5+5 \\ c &= 10\\ \therefore y &= -x + 10 \end{align*}

\begin{align*} m &= 3 \text{ (parallel lines)} \\ y &= mx + c \\ (0) &= (3)(4) + c \\ c &= 12\\ \therefore y &= 3x + 12 \end{align*}

\begin{align*} M(x;y) &= (\frac{x_1 + x_2}{2}; \frac{y_1 + y_2}{2}) \\ M_{GH} &= (\frac{-6+(-2)}{2}; \frac{3 + (-3)}{2}) \\ M_{GH} &= (-4; 0) \\ \\ (y -y_1) &= m(x -x_1) \\ m &= \frac{y_2-y_1}{x_2-x_1} \\ \\ (y - 0) &= \frac{1-0}{2-(-4)}(x - (-4)) \\ \therefore y &= \frac{1}{6}(x + 4) \\ \therefore y &= \frac{1}{6}x + \frac{2}{3} \end{align*}

To calculate the lengths of the sides we need to use the distance formula. The four sides are \(FG\), \(GH\), \(HI\) and \(FI\)

\begin{align*} d_{FG} &= \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\ &= \sqrt{(1 - 2)^2 + (5 - 0)^2}\\ &= \sqrt{(-1)^2 + (5)^2}\\ &= \sqrt{26}\\\\ d_{GH} &= \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\ &= \sqrt{(3 - 1)^2+(7 - 5)^2}\\ &= \sqrt{(2)^2 + (2)^2}\\ &= \sqrt{8}\\\\ d_{HI} &= \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\ &= \sqrt{(7 - 3)^2 + (2 - 7)^2}\\ &= \sqrt{(4)^2 + (-5)^2}\\ &= \sqrt{41} \\\\ d_{FI} &= \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\ &= \sqrt{(2 - 7)^2 + (0 - 2)^2}\\ &= \sqrt{(-5)^2 + (-2)^2}\\ &= \sqrt{29} \end{align*}

We want to know if \(GH \parallel FI\) and \(FG \parallel HI\). We can calculate the gradient of each of the sides and then compare the gradients.

\begin{align*} m_{FG} &= \dfrac{y_2 - y_1}{x_2 - x_1} \\ &= \frac{0 - 5}{2 - 1} \\ &= \frac{-5}{1} \\ &= -5 \\\\ m_{HI} &= \dfrac{y_2 - y_1}{x_2 - x_1} \\ &= \frac{2 - 7}{7 - 3} \\ &= \frac{-5}{4} \\ &= -\frac{5}{4} \\\\ m_{GH} &= \dfrac{y_2 - y_1}{x_2 - x_1} \\ &= \dfrac{5 - 7}{1 - 3} \\ &= \dfrac{-2}{-2} \\ &= 1 \\\\ m_{FI} &= \dfrac{y_2 - y_1}{x_2 - x_1} \\ &= \frac{0 - 2}{2 - 7} \\ &= \frac{-2}{-5} \\ &= \frac{2}{5} \end{align*}

We note that \(m_{FG} \ne m_{HI}\) and \(m_{GH} \ne m_{FI}\) therefore the opposite sides are not parallel.

To determine if the diagonals bisect each other we need to find the mid-point of \(FH\) and \(GI\).

\begin{align*} M_{GI} &= \left(\frac{x_1 + x_2}{2} ; \frac{y_1 + y_2}{2}\right)\\ &= \left(\frac{1 + 7}{2} ; \frac{5 + 2}{2}\right)\\ &= \left(\frac{8}{2} ; \frac{7}{2}\right)\\ &= \left(4 ; \frac{7}{2}\right)\\\\ M_{FH} &= \left(\frac{x_1 + x_2}{2} ; \frac{y_1 + y_2}{2}\right)\\ &= \left(\frac{3 + 2}{2} ; \frac{7 + 0}{2}\right)\\ & = \left(\frac{5}{2} ; \frac{7}{2}\right) \end{align*}

Therefore \(M_{GI} \ne M_{FH}\) and the diagonals do not bisect each other.

This is an ordinary quadrilateral. The opposite sides are not parallel, the diagonals do not bisect each other and none of the sides are equal in length.

To calculate the lengths of the sides we need to use the distance formula. The four sides are \(AB\), \(BC\), \(CD\) and \(AD\)

\begin{align*} d_{AB} &= \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \\ &= \sqrt{(3 - 4)^2 + (2 - 5)^2} \\ &= \sqrt{(-1)^2 + (-3)^2} \\ &= \sqrt{10} \\\\ d_{BC} &= \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \\ &= \sqrt{(4 - 1)^2 + (5 - 7)^2} \\ &= \sqrt{(3)^2 + (-2)^2} \\ &= \sqrt{13} \\\\ d_{CD} &= \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \\ &= \sqrt{(1 - 1)^2 + (7 - 3)^2} \\ &= \sqrt{0 + (4)^2} \\ &= \sqrt{16} \\ &= 4 \\\\ d_{AD} &= \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \\ &= \sqrt{(3 - 1)^2 + (2 - 3)^2} \\ &= \sqrt{(2)^{2} + (-1)^2} \\ &= \sqrt{5} \end{align*}

An isosceles trapezium; one pair of opposite sides equal in length and one pair of opposite sides parallel.

To show this we need to find the mid-points of \(AC\) and \(BD\).

\begin{align*} M_{AC} & = \left(\frac{x_1 + x_2}{2} ; \frac{y_1 + y_2}{2}\right) \\ & = \left(\frac{0 + 5}{2} ; \frac{3 - 1}{2}\right) \\ & = \left(\frac{5}{2} ; \frac{2}{2}\right) \\ & = \left(\frac{5}{2} ; 1\right) \\\\ M_{BD} &= \left(\frac{x_1 + x_2}{2} ; \frac{y_1 + y_2}{2}\right) \\ & = \left(\frac{4 - 1}{2} ; \frac{3 - 1}{2}\right) \\ & = \left(\frac{3}{2} ; \frac{2}{2}\right)\\ & = \left(\frac{3}{2} ; 1\right) \end{align*}

\(M_{AC} \ne M_{BD}\), therefore the diagonals do not bisect each other.

Trapezium. One pair of opposite sides parallel.

First draw a sketch of the quadrilateral:

\begin{align*} m_{PS} &= \frac{y_2 - y_1}{x_2 - x_1}\\ &= \frac{-3 + 2}{0 - 3} \\ & = \frac{-1}{-3} \\ &= \frac{1}{3} \end{align*}

\begin{align*} M_{QR} & = \left(\frac{x_1 + x_2}{2} ; \frac{y_1 + y_2}{2}\right) \\ & = \left(\frac{0 + 3}{2} ; \frac{-3 + 2}{2}\right) \\ & = \left(\frac{3}{2} ; \frac{-1}{2}\right) \end{align*}

We need to calculate the gradients of each of the sides to see if opposite sides are parallel. We have calculated the gradient of \(PS\) so we only need to check the gradients of the other three sides. However, looking at our sketch \(PS\) is not parallel to \(QR\) (you can check this by calculating the gradient of \(QR\)).

\begin{align*} m_{RS} &= \frac{y_2 - y_1}{x_2 - x_1} \\ &= \frac{-2 - 2}{3 - 3} \\ &= \frac{-4}{0} \\ & = \text{undefined} \end{align*}

And,

\begin{align*} m_{QR} &= \frac{y_2 - y_1}{x_2 - x_1} \\ &= \frac{2 - 5}{3 - (-2)} \\ & = \frac{-3}{5} \end{align*}

Therefore \(PQRS\) is not a parallelogram. Opposite sides are not parallel.

We need to show that two sides are equal in length. We therefore calculate the length of each of the sides of the triangle.

\begin{align*} d_{AB} &= \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \\ &= \sqrt{(1 - 4)^2 + (3 - 1)^2} \\ & = \sqrt{(-3)^2 + (2)^2} \\ & = \sqrt{13} \\\\ d_{BC} &= \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \\ &= \sqrt{(4 - 6)^2 + (1 - 4)^2} \\ &= \sqrt{(-2)^2 + (-3)^2} \\ &= \sqrt{13} \\\\ d_{AC} &= \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \\ &= \sqrt{(1 - 6)^2 + (3 - 4)^2} \\ & = \sqrt{(-5)^2 + (-1)^2} \\ &= \sqrt{26} \end{align*}

Two sides of the triangle are equal in length, therefore \(\triangle ABC\) is isosceles.

\begin{align*} M_{AC} & = \left(\frac{x_1 + x_2}{2} ; \frac{y_1 + y_2}{2}\right)\\ & = \left(\frac{1 + 6}{2} ; \frac{3 + 4}{2}\right)\\ & = \left(\frac{7}{2} ; \frac{7}{2}\right) \end{align*}

\begin{align*} m_{AB} &= \frac{y_2 - y_1}{x_2 - x_1}\\ &= \frac{1 - 3}{4 - 1}\\ & = \frac{-2}{3} \end{align*}

We have just calculated the gradient of \(AB\): \(m_{AB} = \dfrac{-2}{3}\). We need to calculate the gradient of \(BD\) and \(AD\):

\begin{align*} m_{BD}& = \frac{y_2 - y_1}{x_2 - x_1} \\ &= \frac{-1 - 1}{7 - 4} \\ & = \frac{-2}{3} \\\\ m_{AD} &= \frac{y_2 - y_1}{x_2 - x_1} \\ &= \frac{-1 - 3}{7 - 1} \\ & = \frac{-2}{3} \\ \\ m_{AB} & = m_{BD} = m_{AD} \end{align*}

Therefore \(A\), \(B\) and \(D\) are collinear.

Therefore \(D\) lies on line \(AB\).

Any two of the following:

• Both pairs of opposite sides are parallel.
• Both pairs of opposite sides are equal in length.
• Both pairs of opposite angles are equal.
• Both diagonals bisect each other.

We need to show that both pairs of opposite sides are parallel. So we need to calculate the gradient of each of the sides:

\begin{align*} m_{AB} & = \frac{1 + 4}{-4 + 2} \\ & = \frac{5}{-2} \\ m_{CD} & = \frac{-1 - 4}{4 - 2} \\ & = \frac{-5}{2} \\ \therefore AB \parallel CD \\\\ m_{BC} & = \frac{4 - 1}{2 + 4} \\ & = \frac{3}{6} = \frac{1}{2} \\ m_{AD} & = \frac{-1 + 4}{4 + 2} \\ & = \frac{3}{6} = \frac{1}{2} \\ \therefore BC \parallel AD \end{align*}

Therefore \(ABCD\) is a parallelogram (2 pairs opp. sides \(\parallel\)).

Since the diagonals bisect each other we know that \(AC \perp BD\). Therefore:

\begin{align*} m_{AC} \times m_{BD} & = -1 \\ m_{AC} \times \frac{6 - 2}{2 - 4} & = -1 \\ m_{AC} \times -2 & = -1 \\ m_{AC} & = \frac{1}{2} \end{align*}

From above we have that \(m = \frac{1}{2}\).

Substitute point \(A\) into the equation:

\begin{align*} (0) & = \frac{1}{2}(-5) + c \\ 0 & = \frac{-5}{2} + c \\ \therefore c & = \frac{5}{2} \\ \therefore y & = \frac{1}{2}x + \frac{5}{2} \\ \therefore 2y & = x + 5 \end{align*}

We first calculate the coordinates of \(E\):

\begin{align*} E(\frac{2 + 4}{2};\frac{6 + 2}{2}) \\ E(3;4) \end{align*}

Now we can use the coordinates of \(E\) to find the coordinates of \(C\). Since \(E\) is the mid-point of \(AC\) we can use the mid-point formula to find \(C\):

\begin{align*} \frac{-5 + x}{2} & = 3 \\ -5 + x & = 6 \\ x & = 11 \\\\ \frac{0 + y}{2} & = 4 \\ y & = 8 \end{align*}

Therefore \(C(11;8)\).

\begin{align*} BC & = \sqrt{(-2+6)^{2} + (3+3)^{2}} \\ & = \sqrt{4^2 + 9^2} \\ & = \sqrt{16 + 81} \\ & = \text{9,8} \end{align*}

\begin{align*} m_{AC} & = \frac{3 + 1}{-2 - 4} \\ & = \frac{4}{-6} \\ & = \frac{-2}{3} \end{align*}

\begin{align*} m_{AP} & = \frac{19 + 1}{-26 - 4} \\ & = \frac{20}{-30} \\ & = \frac{-2}{3} \end{align*}

From the previous question we have that \(m_{AC} = \frac{-2}{3}\), therefore \(m_{AC} = m_{AP}\).

Therefore \(A\), \(C\) and \(P\) are collinear.

\begin{align*} m_{BC} & = \frac{3 + 3}{-2 + 6} \\ & = \frac{6}{4} \\ & = \frac{3}{2} \\ \therefore y & = \frac{3}{2}x + c \end{align*}

Substitute \(B(-6;-3)\):

\begin{align*} (-3) & = \frac{3}{2}(-6) + c \\ -3 & = -9 + c \\ \therefore c & = 6 \\ \therefore y & = \frac{3}{2}x + 6 \end{align*}

The equation of line \(BC\) is \(y = \dfrac{3}{2}x + 6\).

For a triangle to be right-angled the gradients of two of the sides must be perpendicular. We need to calculate \(m_{AC}\), \(m_{BC}\) and \(m_{AB}\).

We will start with \(m_{AC}\) and \(m_{BC}\) since we have these values from previous questions: \(m_{AC} = \dfrac{-2}{3}\) and \(m_{BC} = \dfrac{3}{2}\).

We can check if these two gradients are perpendicular:

\begin{align*} m_{AC} \times m_{BC} & = \frac{-2}{3} \times \frac{3}{2} \\ & = \frac{-6}{6} \\ & = -1 \end{align*}

Since the product of these two gradients is \(-\text{1}\), \(AC \perp BC\).

Therefore \(\triangle ABC\) is right-angled.

\begin{align*} M_{AB} & = \left(\frac{-2 + 4}{2}; \frac{2 + 5}{2}\right) \\ & = \left(1 ; \frac{7}{2}\right) \end{align*}

Therefore \(a = 1\) and \(b = \frac{7}{2}\).

Let the line be \(FG\).

First calculate the gradient of \(BC\):

\begin{align*} m_{BC} & = \frac{2 - 5}{8 - 4} \\ m_{BC} & = \frac{-3}{4} \end{align*}

Now we can calculate the gradient of \(FG\):

\begin{align*} -1 & = m_{BC} \times m_{FG} \\ -1 & = \left(\frac{-3}{4}\right)m_{FG} \\ m_{FG} & = \frac{4}{3} \end{align*}

Therefore we have \(y = \dfrac{4}{3}x - c\). Since the line passes through the origin the \(y\)-intercept is \(\text{0}\).

Therefore the equation of the line perpendicular to \(BC\), which passes through the origin is: \(y = \dfrac{4}{3}x\).

\begin{align*} M_{BD} & = \left(\frac{4 + 1}{2}; \frac{5 - 3}{2}\right) \\ & = \left(\frac{7}{2} ; 1\right) \end{align*}

\begin{align*} M_{AC} & = \left(\frac{-2 + 8}{2}; \frac{2 + 2}{2}\right) \\ & = \left(3 ; 2\right) \end{align*}

The mid-point of \(BD\) is not the mid-point of \(AC\). Therefore the diagonals of the quadrilateral do not bisect each other and \(ABCD\) is not a parallelogram.

We use the mid-point of \(BD\) to find the new \(x\) and \(y\) coordinates of \(C\):

\begin{align*} \frac{5}{2} & = \frac{-2 + x}{2} \\ \frac{10}{2} & = -2 + x \\ x & = 7 \\\\ 1 & = \frac{2 + y}{2} \\ 2 & = 2 + y \\ 0 & = y \end{align*}

Therefore \(C(7;0)\) would make \(ABCD\) a parallelogram.

\begin{align*} m_{AB} & = \frac{4 - 7}{8 + 1} \\ & = \frac{-3}{9} \\ & = -\frac{1}{3} \end{align*}

First draw a sketch:

For the triangle to be right-angled at \(B\), \(m_{AB} \times m_{BC} = -1\). We have \(m_{AB}\) from the previous question.

\begin{align*} m_{BC} & = \frac{-5 - 4}{5 - 8} \\ & = \frac{-9}{-3} \\ & = 3 \\\\ \therefore m_{AB} \times m_{BC} & = 3 \times \frac{-1}{3} \\ & = -1 \end{align*}

Therefore \(BC \perp AB\) and \(\triangle ABC\) is right-angled at \(B\).

\begin{align*} AB & = \sqrt{(8 - (-1))^{2} + (4 - 7)^{2}} \\ & = \sqrt{90} \\ & = \text{9,49} \end{align*}

First find the mid-point of \(BC\):

\begin{align*} M_{BC} & = \left(\frac{8 + 5}{2}; \frac{4 - 5}{2}\right) \\ & = \left(\frac{13}{2}; \frac{-1}{2}\right) \end{align*}

Now we can calculate the gradient of the line:

\begin{align*} m & = \frac{7 - \frac{-1}{2}}{-1 - \frac{13}{2}} \\ & = -1 \\ \therefore y & = -x + c \end{align*}

Substitute \(A(-2;2)\):

\begin{align*} (2) & = -(-1) + c \\ c & = 6 \\ \therefore y & = -x + 6 \end{align*}

The equation of the the line from \(A\) to the mid-point of \(BC\) is \(y = -x + 6\).

First find the length of \(BC\):

\begin{align*} BC & = \sqrt{(8-5)^{2} + (4+5)^{2}} \\ & = 3\sqrt{10} \\ & = \text{9,5} \end{align*}

Now we can find the area:

\begin{align*} \text{Area } & = \frac{1}{2}bh \\ & = \frac{1}{2}(AB)(BC) \\ & = \frac{1}{2}(3\sqrt{10})(3\sqrt{10}) \\ & = \text{45} \end{align*}

We first draw a sketch. We note that \(D\) lies somewhere on the line \(x = 4\).

For parallel lines \(m_{AD} = m_{CD}\):

\begin{align*} \frac{k - 5}{4 - 0} & = \frac{k + 5}{4 - 0} \\ \frac{k - 5}{4} & = \frac{k + 5}{4} \\ \therefore k - 5 & = k + 5 \\ 0 & = 10 \end{align*}

Therefore there is no value of \(k\) such that \(AD\) will be parallel to \(DC\).

We can see that this must be true since \(D\) is a common point on lines \(AD\) and \(CD\) so \(AD\) cannot be parallel to \(CD\).

\begin{align*} \sqrt{52} & = \sqrt{(4-0)^{2} + (k + 5)^{2}} \\ \sqrt{52} & = \sqrt{16 + (k + 5)^{2}} \\ \therefore 52 & = 16 + k^{2} + 10k + 25 \\ 0 & = k^{2} + 10k - 11 \\ & = (k + 11)(k - 1) \\ \therefore k = -11 & \text{ or } k = 1 \end{align*}

But \(k \geq 0\), therefore \(k = 1\)

\begin{align*} PQ & = \sqrt{(-3 - 7)^{2} + (-1 - 4)^{2}} \\ & = 5\sqrt{5} \end{align*}

\begin{align*} m_{PQ} & = \frac{-1 - 4}{7 + 3} \\ & = \frac{-5}{10} \\ & = -\frac{1}{2} \end{align*}

\(y = -\frac{1}{2}x + c\). Substitute \(P(-3;4)\):

\begin{align*} (4) & = -\frac{1}{2}(-3) + c \\ c & = \frac{5}{2} \\ \therefore y & = -\frac{1}{2}x + \frac{5}{2} \end{align*}

The equation of the \(PQ\) is \(y = -\frac{1}{2} + \frac{5}{2}\).

\begin{align*} m_{QR} & = m_{PQ} \\ \frac{b + 1}{3 - 7} & = -\frac{1}{2} \\ b + 1 & = 2 \\ b & = 1 \end{align*}

\begin{align*} M\left(\frac{4 + (-2)}{2}; \frac{9-3}{2}\right) \\ M(1;3) \end{align*}

\begin{align*} m_{AB} & = \frac{-3 - 9}{-2 -4} \\ & = \frac{-12}{-6} \\ & = 2 \end{align*}

\begin{align*} m & = -1 \div m_{AB} \\ & = -1 \div 2 \\ & = -\frac{1}{2} \end{align*}

From the previous question we have that \(y = -\frac{1}{2}x + c\). Substitute \(M\):

\begin{align*} 3 & = -\frac{1}{2}(1) + c \\ c & = \frac{7}{2} \\ \therefore y & = -\frac{1}{2}x + \frac{7}{2} \end{align*}

The equation of the perpendicular bisector of \(AB\) is \(y = -\frac{1}{2}x + \frac{7}{2}\)

Since the line is parallel to \(AB\) the gradient is the same and so we have: \(y = 2x + c\). Substitute: \((0;6)\):

\begin{align*} (6) & = 2(0) + c \\ c & = 6 \\ \therefore y & = 2x + 6 \end{align*}

The equation of the line parallel to \(AB\) and passing through \((0;6)\) is \(y = 2x + 6\).

Since \(LMNP\) is a parallelogram we know that \(LM \parallel NP\). Therefore:

\begin{align*} m_{LM} & = m_{NP} \\ \frac{4 + 1}{-2 + 1} & = \frac{y - 0}{x -4} \\ \frac{5}{-1} & = \frac{y}{x - 4} \\ \therefore y & = 5 \\ -1 & = x - 4 \\ 3 & = x \end{align*}

Therefore \(N(3;5)\).

We can now sketch the parallelogram:

\begin{align*} m_{MP} & = \frac{0 - 4}{4 + 2} \\ & = \frac{-4}{6} \\ & = \frac{-2}{3} \\\\ m_{LN} & = \frac{5 + 1}{3 + 1} \\ & = \frac{6}{4} \\ & = \frac{3}{2} \\\\ m_{MP} \times m_{LN} & = \frac{-2}{3} \times \frac{3}{2} \\ & = -1 \\ \therefore MP \perp LN \end{align*}

\(LMNP\) is a rhombus since the diagonals intersect at right-angles.

We can calculate the length of each side and show that all four lengths are the same:

\begin{align*} MN & = \sqrt{(-2 - 3)^{2} + (4 - 5)^{2}} \\ & = \sqrt{26} \\\\ NP & = \sqrt{(3 - 4)^{2} + (5 - 0)^{2}} \\ & = \sqrt{26} \\\\ LP & = \sqrt{(-1 - 4)^{2} + (-1 - 0)^{2}} \\ & = \sqrt{26} \\\\ ML & = \sqrt{(-2 + 1)^{2} + (4 + 1)^{2}} \\ & = \sqrt{26} \end{align*}

Therefore \(LMNP\) is a square, all the sides are equal in length.

\begin{align*} m_{BC} & = \frac{0 + 2}{4 + 4} \\ & = \frac{2}{8} \\ & = \frac{1}{4} \end{align*}

\begin{align*} D & = \left(\frac{-2 - 4}{2} ; \frac{-2 + 4}{2}\right) \\ & = (-3;1) \end{align*}

\begin{align*} DE & = \sqrt{(1-(-3))^{2} + (2-1)^{2}} \\ & = \sqrt{17} \\ & = \text{4,1} \end{align*}

\begin{align*} m_{DE} & = \frac{1 - 2}{-3 - 1} \\ & = \frac{1}{4} \end{align*}

A conjecture regarding lines \(BC\) and \(DE\) is \(DE \parallel BC\).

From earlier we have the gradient of \(BC\), so the equation of \(BC\) is: \(y = \frac{1}{4}x + c\). Substitute \(B\):

\begin{align*} (-2) & = \frac{1}{4}(-4) + c \\ c & = -1 \\ y & = \frac{1}{4}x - 1 \end{align*}

The equation of \(BC\) is \(y = -\frac{1}{2}x - 4\).

\begin{align*} PQ & = \sqrt{(-3 - 1)^{2} + (3 + 2)^{2}} \\ & = \sqrt{41} \end{align*}

\(M\) is the mid-point of both \(PR\) and \(QS\) (\(PQRS\) is a parallelogram). Since we do not know the coordinates of \(S\) we will use \(PR\) to find \(M\).

\begin{align*} M\left(\frac{-3 + 5}{2}; \frac{3 + 1}{2}\right) \\ \therefore M(1;2) \end{align*}

\begin{align*} T\left(\frac{-3 + 1}{2}; \frac{3 - 2}{2}\right) \\ \therefore T\left(-1;\frac{1}{2}\right) \end{align*}

\begin{align*} m_{MT} & = \frac{2 - \frac{1}{2}}{1 + 1} \\ & = \frac{3}{4} \\\\ m_{QR} & = \frac{-2 - 1}{1 - 4} \\ & = \frac{3}{4} \\\\ \therefore m_{MT} & = m_{QR} \\ \therefore MT \parallel QR \end{align*}

We can use the coordinates of \(M\) to find the coordinates of \(S\):

\[M(1;2) = S\left(\frac{x + 1}{2}; \frac{y - 2}{2}\right)\]

Solve for \(x\):

\begin{align*} 1 & = \frac{x + 1}{2} \\ 2 & = x + 1 \\ x & = 1 \end{align*}

Solve for \(y\):

\begin{align*} 2 & = \frac{y - 2}{2} \\ 4 & = y - 2 \\ y & = 6 \end{align*}

Therefore \(S(1;6)\).

\begin{align*} PQ & = \sqrt{(5 - 1)^{2} + (1 - 3)^{2}} \\ & = 2\sqrt{5} \\\\ QR & = \sqrt{(1 - 1)^{2} + (3 + 2)^{2}} \\ & = 5 \\\\ PR & = \sqrt{(5 - 1)^{2} + (1 + 2)^{2}} \\ & = 5 \\\\ \therefore QR & = PR \text{ triangle is isosceles} \end{align*}

\(\triangle PQR\) is an isosceles triangle.

\begin{align*} S\left(\frac{5+1}{2}; \frac{1+3}{2}\right) \\ \therefore S(3;2) \\\\ T\left(\frac{1+1}{2}; \frac{3-2}{2}\right) \\ \therefore T\left(1;\frac{1}{2}\right) \end{align*}

\begin{align*} m_{ST} & = \frac{2 - \frac{1}{2}}{3 - 1} \\ & = \frac{\frac{3}{2}}{2} \\ & = \frac{3}{4} \end{align*}

\begin{align*} m_{PR} & = \frac{1 + 2}{5 - 1} \\ & = \frac{3}{4} \\ \therefore ST \parallel PR & \text{ (equal gradients)} \end{align*}

\(Q\) lies on the \(y\)-axis so we have \(Q(0;y)\). Using the equation of \(QR\) we get:

\begin{align*} -3y & = -6 \\ \therefore y & = 2 \\ \therefore Q(0;2) \end{align*}

\begin{align*} y & = \frac{1}{3}x + 2 \\ \therefore m_{QR} & = \frac{1}{3} \\\\ m_{QP} & = -3 \\\\ m_{QP} \times m_{QR} & = \frac{1}{3} \times -3 \\ & = -1 \end{align*}

Therefore \(PQ \perp QR\).

The equation of \(PR\) is \(x - y - 2 = 0\). In standard form this is \(y = x - 2\).

Therefore \(m_{PR} = 1\).

We use the equation of \(PR\) and substitute in \(y = 4\):

\begin{align*} x - 4 - 2 & = 0 \\ \therefore x & = 6 \end{align*}

We first calculate \(S\):

\begin{align*} S\left(\frac{0 + 6}{2} ; \frac{2 + 4}{2}\right) \\ S\left(3; 3\right) \end{align*}

Now we can find the equation of the line \(SP\):

\begin{align*} m_{SP} & = \frac{-1 - 3}{1 - 3} \\ & = \frac{-4}{-2} \\ & = 2 \\ \therefore y & = 2x + c \\ -1 & = 2(1) + c \\ y & = 2x - 3 \end{align*}

The equation of the line from \(P\) to \(S\) is \(y = 2x - 3\).

Since \(E\), \(L\) and \(S\) are collinear, \(m_{EL} = m_{LS}\). Therefore:

\begin{align*} m_{EL} & = \frac{5}{6} \\ \therefore \frac{5 - \text{2,5}}{3 - (t + 1)} & = \frac{5}{6} \\ \frac{\text{2,5}}{2 - t} & = \frac{5}{6} \\ 6(\text{2,5}) & = 5(2 - t) \\ 15 & = 10 - 5t \\ 5t & = -5 \\ t & = -1 \end{align*}

\begin{align*} y & = \frac{5}{6}x + c \\ \therefore 0 & = \frac{5}{6}(-3) + c \\ c & = \frac{5}{2} \\ y & = \frac{5}{6}x + \frac{5}{2} \\ \end{align*}

Now we can rearrange the equation:

\begin{align*} \frac{5}{2} & = y - \frac{5}{6}x \\ 1 & = \frac{2}{5}y - \frac{1}{3}x \\ & = \frac{2y}{5} - \frac{x}{3} \end{align*}

We have \(\dfrac{x}{a} = -\dfrac{x}{3}\), therefore \(a = -3\).

We have \(\dfrac{y}{b} = \frac{2y}{5}\), therefore \(b = \dfrac{5}{2}\).

\begin{align*} d &= \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2} \\ AC &= \sqrt{(2-(-3))^2 + (-5-(-4))^2} \\ &= \sqrt{5^2 (-1)^2} \\ &= \sqrt{25+1} \\ &= \sqrt{26} \\ \\ d &= \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2} \\ BD &= \sqrt{(0-(-1))^2 + (-2-(-7))^2} \\ &= \sqrt{1^2 + 5^2} \\ &=\sqrt{26} \end{align*}

Therefore \(d_{AC} = d_{BD} =\sqrt{26}\).

\begin{align*} M(x;y) & = \left(\frac{x_1 + x_2}{2}; \frac{y_1 + y_2}{2}\right) \\ & = \left(\frac{-1 +0}{2}; \frac{-7 + (-2)}{2}\right) \\ & = \left(\frac{-1}{2}; \frac{-9}{2}\right) \\ M &= (-\text{0,5};-\text{4,5}) \end{align*}

\begin{align*} m &= \frac{y_2 - y_1}{x_2 - x_1} \\ m_{AM} &= \frac{-4 - (-\text{4,5})}{-3 - (-\text{0,5})} \\ & = \frac{\text{0,5}}{-\text{2,5}} \\ & = -\frac{1}{5} \\ \\ m_{BD} & = \frac{-7 - (-2)}{-1 - (0)} \\ &= \frac{-5}{-1} \\ &= 5 \\ \\ m_{BD} \times m_{AM} &= -\frac{1}{5} \times 5 \\ &= -1 \end{align*}

Therefore \(AM \perp BD\).

\begin{align*} m_{MC} &= \frac{-5 - (-\text{4,5})}{-2 - (-\text{0,5})} \\ &= \frac{-\text{0,5}}{\text{2,5}} \\ & = -\frac{1}{5} \end{align*}

From earlier we know that \(m_{AM} = -\dfrac{1}{5}\).

\(m_{MC} = m_{AM}\) and \(M\) is a common point.

Therefore \(A\), \(M\) and \(C\) are collinear.

First draw a sketch:

From earlier we found that \(d_{AC} = d_{BD} =\sqrt{26}\). Therefore the diagonals are equal in length.

Since the diagonals are equal in length we know that the quadrilateral must be a square.

We can confirm this by showing that all four sides are equal in length and that \(AB \perp BC\), \(BC \perp CD\), \(CD \perp AD\) and \(AD \perp AB\).

\begin{align*} M(x;y) &= (\frac{x_1 + x_2}{2}; \frac{y_1 + y_2}{2}) \\ (2;-2)&= (\frac{3 + x_B}{2}; \frac{1 + y_B}{2}) \\ \\ 2 &= \frac{3 + x_B}{2} \\ 4 &= 3 + x_B \\ x_B & = 1 \\ \\ -2 &= \frac{1 + y_B}{2} \\ -4 &= 1 + y_B \\ y_B &= -5 \\ \\ \therefore B(x;y) &= (1;-5) \end{align*}

\begin{align*} m &= \frac{y_2-y_1}{x_2-x_1} \\ m_{AM} &= \frac{-2-1}{2-3} \\ &= \frac{-3}{-1} \\ &= 3 \end{align*}

\begin{align*} y &= mx + c \\ (-2) &= (3)(2) + c \\ c &= -2 - 6 \\ c &= -8 \\ \\ \therefore y &= 3x -8 \end{align*}

The perpendicular bisector of \(AB\) passes through \(M\) and has a gradient \(-\dfrac{1}{m_{AB}}\).

\begin{align*} m &= -\frac{1}{m_{AB}} \\ m &= -\frac{1}{3} \\ \\ y &= mx + c \\ (-2) &= (-\frac{1}{3})(2) + c \\ c &= \frac{2}{3} -2 \\ c &= -\frac{4}{3}\\ \therefore y &= -\frac{1}{3}x -\frac{4}{3} \end{align*}

\begin{align*} M(x;y) &= \left(\frac{x_1 + x_2}{2}; \frac{y_1 + y_2}{2}\right) \\ E &= \left(\frac{5 + (-4)}{2}; \frac{0 + (-3)}{2}\right) \\ \therefore E &= \left(\frac{1}{2}; -\frac{3}{2}\right) \end{align*}

\begin{align*} M(x;y) &= \left(\frac{x_1 + x_2}{2}; \frac{y_1 + y_2}{2}\right) \\ M_{AC} &= \left(\frac{-3 + 4}{2}; \frac{6 + (-9)}{2}\right) \\ &= \left(\frac{1}{2}; -\frac{3}{2}\right) \\ \therefore M_{AC} &= E \end{align*}

\(ABCD\) is a parallelogram (Diagonals bisect at \(E\)).

\begin{align*} (y -y_1) &= m(x -x_1) \\ m &= \frac{y_2-y_1}{x_2-x_1} \\ \\ (y - 0) &= \frac{-3-0}{-4-(5)}(x - (5)) \\ \therefore y &= \frac{-3}{-9}(x - 5) \\ \therefore y &= \frac{1}{3}x - \frac{5}{3} \end{align*}

The perpendicular bisector of \(BD\) passes through \(E\) and has a gradient \(-\dfrac{1}{m_{BD}}\)

\begin{align*} m &= -\frac{1}{m_{BD}} \\ m &= \frac{1}{-\frac{1}{3}} \\ m &= -3 \\ \\ y &= mx + c \\ - \frac{3}{2} &= (3)(\frac{1}{2}) + c \\ c &= - \frac{3}{2} + \frac{3}{2} \\ c &= 0\\ \therefore y &= -3x \end{align*}

\begin{align*} m &= \frac{y_2 - y_1}{x_2 - x_1} \\ &= \frac{6 - (-9)}{-3 - 4} \\ m_{AC} &= -\frac{15}{7} \end{align*}

\begin{align*} m_{AC} \times m_{BD} &= -\frac{15}{7} \times \frac{1}{3} \\ &= -\frac{15}{21} \neq -1 \end{align*}

No, \(ABCD\) is not a rhombus because the diagonals do not intersect at right angles.

\begin{align*} d &= \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \\ AB &= \sqrt{(-3 - 5))^2 + (6 - 0)^2} \\ &= \sqrt{8^2 + 6^2} \\ &= \sqrt{64 + 36} \\ &= \sqrt{100} \\ &= 10 \end{align*}

We can substitute \(U\) into \(f(x)\) to find \(a\):

\begin{align*} f(x) &= \frac{3}{2}x - 4 \\ a &= \frac{3}{2}(6) - 4 \\ &= 9 - 5 \\ \therefore a &= 5 \end{align*}

We know that \(m_{g(x)} \times m_{f(x)} = -1\), therefore \(m_{g(x)} = -\dfrac{2}{3}\).

\begin{align*} g(x) &= -\frac{2}{3}x + c \\ 5 &= -\frac{2}{3}(6) + c \\ c &= 5 + 4 \\ c &= 9 \end{align*}

Therefore \(g(x) = -\dfrac{2}{3}x + 9\). Substitute in \(V\) to solve for \(b\):

\begin{align*} 4 &= -\frac{2}{3}b + 9 \\ -5 &= -\frac{2}{3}b \\ b &= \frac{15}{2} \\ \therefore b &= 7\frac{1}{2} \end{align*}

\(U\), \(V\) and \(W\) are collinear and the equation for the line is \(g(x) = -\frac{2}{3}x + 9\). We can substitute \(W\) into the equation for the line to solve for \(c\):

\begin{align*} c &= -\frac{2}{3}(1) + 9 \\ \therefore c &= 8\frac{1}{3} \end{align*}

\begin{align*} m &= \frac{y_2 - y_1}{x_2 - x_1} \\ &= \frac{4 - 2}{-1 - 5} \\ m_{MN} &= \frac{2}{-6} \\ m_{MN} &= -\frac{1}{3} \end{align*}

\begin{align*} y &= mx + c \\ 4 &= -\frac{1}{3}(-1) + c \\ c &= 4 -\frac{1}{3} \\ c &= 3\frac{2}{3} \\ y &= -\frac{1}{3}x + 3\frac{2}{3} \end{align*}

First find the coordinates of \(A\):

\begin{align*} M_{OA} &= \left(\frac{x_1 + x_2}{2}; \frac{y_1 + y_2}{2}\right) \\ (-1,4) &= \left(\frac{x_A + 0}{2}; \frac{y_A + 0}{2}\right) \\ \\ -1 &= \frac{x_A}{2} \\ x_A &= -2 \\ 4 &= \frac{y_A}{2} \\ y_A &= 8 \\ \therefore A(x;y) &= (-2;8) \end{align*}

Next find the coordinates of \(B\):

\begin{align*} M_{OB} &= \left(\frac{x_1 + x_2}{2}; \frac{y_1 + y_2}{2}\right) \\ (5,2) &= \left(\frac{x_B }{2}; \frac{y_B}{2}\right) \\ \\ 5 &= \frac{x_B}{2} \\ x_B &= 10 \\ 2 &= \frac{y_B}{2} \\ y_B &= 4 \\ \therefore B(x;y) &= (10;4) \end{align*}

Now we can calculate the gradient of \(AB\) and compare it to the gradient of \(MN\):

\begin{align*} m &= \frac{y_2-y_1}{x_2-x_1} \\ m_{AB} &= \frac{8-4}{-2-10} \\ &= -\frac{4}{12} \\ m_{AB} &= -\frac{1}{3} = m_{MN}\\ &\therefore AB \parallel MN \end{align*}

We note the following:

• \(OA = 2OM\) and \(OB = 2ON\) (Mid-points)
• \(AB = 2MN\) (Mid-point theorem)

From this we can see that \(\triangle OAB\) is twice the size of \(\triangle OMN\) (each side of \(\triangle OAB\) is twice the size of the same side in \(\triangle OMN\)). Therefore the area of \(\triangle OAB\) is twice the area of \(\triangle OMN\):

\begin{align*} \frac{\text{area } \triangle OAB}{\text{area } \triangle OMN} & = \frac{2 \text{area } \triangle OMN}{\text{area } \triangle OMN} \\ = 2 \end{align*}

We can use the fact that the diagonals of a parallelogram bisect each other to find \(P\). The mid-point of \(AB\) must be the same as the mid-point of \(OP\).

\begin{align*} M_{AB} &= \left(\frac{x_1 + x_2}{2}; \frac{y_1 + y_2}{2}\right) \\ M_{AB} &= \left(\frac{-2 + 10}{2}; \frac{8 + 4}{2}\right) \\ M_{AB} &= (4;6) \\ \\ (4;6) &= \left(\frac{x_p + 0}{2}; \frac{y_p + 0}{2}\right) \\ (8;12) &= \left(\frac{x_p}{2}; \frac{y_p}{2}\right) \\ \therefore P(x;y) &= (8;12) \end{align*}

First find the gradient of \(AB\):

\begin{align*} m &= \frac{y_2 - y_1}{x_2 - x_1} \\ &= \frac{-4 - 2}{6 - 8}\\ &= \frac{-6}{-2} \\ & = 3 \end{align*}

Therefore we have \(y = 3x + c\). Now substitute in \(A\):

\begin{align*} -4 &= 3(6) + c \\ c &= -22 \end{align*}

Now we can substitute in \(C\) to solve for \(a\):

\begin{align*} a &= 3(3) - 22 \\ \therefore a &= -13 \end{align*}

\begin{align*} M_{AD} &= \left(\frac{x_1 + x_2}{2}; \frac{y_1 + y_2}{2}\right) \\ (8;2) &= \left(\frac{6 + b}{2}; \frac{-4 + c}{2}\right) \\ \\ 8 &= \frac{6 + b}{2} \\ 16 &= 6 + b \\ b &= 10 \\ \\ 2 &= \frac{-4 + c}{2} \\ 4 &= -4 + c \\ c &= 8 \\ B(x;y) & = (8;12) \end{align*}